When an interfering substance M is present, a fluoride ion selective electrode obeys the following modified response equation:
E(V) = K - ß(0.0592)log{[F-]+Ks[M-]}
A potential of -235.6 mV is measured for a fluoride ion selective electrode in a solution of 3.16×10-5 M fluoride at pH 5.00. Calculate the potential that would be measured if the solution pH were 9.70, assuming the selectivity coefficient (Ks) for OH- is 0.1 and that ß is 1.00. Please show your work.